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Dr.Khaled Kh. Sharaf
Faculty Of Computers And Information
Technology
Second Term 2018 - 2019
DIGITAL DESIGN
Discussion
Chapter 2
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Chapter 2
Chapter 2
Question 1Demonstrate the validity of the following identities by means of truth tables:(a)DeMorgan’s theorem for three variables:
(x + y + z)’ = x’y’z’
x’y’z’z’y’x’(x+y+z)’x+y+zxyz
111110000
001101001
010101010
000101011
011001100
001001101
010001110
000001111
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Chapter 2
Question 1Demonstrate the validity of the following identities by means of truth tables:(a)DeMorgan’s theorem for three variables:
(xyz)’ = x’ + y’ + z’
X +’y +’z’z’y’x’(x+y+z)’x+y+zxyz
111110000
001101001
010101010
000101011
011001100
001001101
010001110
000001111
Chapter 2
Question 1(b) The distributive law: x + yz = ( x + y )( x + z )
(x+y)(x+z)(x+z)(x+y)x+yzxyz
0000000
0100001
0010010
1111011
1111100
1111101
1111110
1111111
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Chapter 2
Question 1Demonstrate the validity of the following identities by means of truth tables:(c) The distributive law: x(y + z) = xy + xz
xy+xzxzxyx(y+z)xyz
0000000
0000001
0000010
0000011
0000100
1101101
1011110
1111111
Chapter 2
Question 1Demonstrate the validity of the following identities by means of truth tables:(d) The associative law: x + (y + z) = (x + y) + z
(x + y) + z(x + y) x+(y+z)(y+z)xxyz
00000000
10110001
11110010
11110011
11101100
11111101
11111110
11111111
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Chapter 2
Question 1Demonstrate the validity of the following identities by means of truth tables:(e) The associative law and x(yz) = (xy)z
(xy)z xyx(yz)yzxyz
0000000
0000001
0000010
0001011
0000100
0000101
0100110
1111111
Chapter 2
Question 2Simplify the following Boolean expressions to a minimum number of literals and draw the original and simplified expressions
xy + xy’ = x(x+y’)= x
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Chapter 2
Question 2
(b)(x + y)(x + y’)= x+yy’=x(x+y’) + y(x+y’)
= xx + xy’ + xy + yy’= x + x(y’+ y) + 0= x + x
= x
remembre xx = xyy’=0y+y’=1X+x=x
Chapter 2
Question 2
(c) xyz + x’y + xyz’ = xy(z+z’) + x’y= xy + x’y= y(x + x’)= y
remembre z+z’=1X+x’=1
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Chapter 2
Question 2
(d) (A + B)’(A’ + B’)’= (A’B’)(AB)=(A’B’)(BA)= A’(B’B)A=0
remembre BB’=0
Chapter 2
Question 2
(e) (a + b + c’)(a’b’ + c)= aa’b’ + ac + ba’b’ + bc + c’a’b’ + c’c= 0 + ac + 0 + bc + c’a’b’ + 0 = ac + bc + a’b’c’
??remembre aa’=0bb’=0
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Chapter 2
Question 2
(f) a’bc + abc’ + abc + a’bc’= a’b(c+c’) + ab(c+c’) = a’b + ab = (a’+ a)b= b
??
remembre a’+ a=1
Chapter 2
Question 3
Find the complement of the following expressions:(a) F = xy’+ x’y
F’ = (xy’+ x’y)’ = (x’ + y)(x + y’)= x’x + x’y’ + xy + yy’= 0 + x’y’ + xy + 0= x’y’ + xy
(b) F = (a + c)(a + b’)(a’ + b + c’)F’ = [(a + c)(a + b’)(a’ + b + c’)]’
= (a+c)’+ (a + b’)’ + (a’+b+c’)’= a’c’ + a’b + ab’c=
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Chapter 2
Question 4
Answer
Chapter 2
Question 5
List the truth table of the function:(a) F = xy + xy’ + y’z (b) F= bc + a’c’
11 10 01 11 00
Fxyzm
00000
10011
00102
00113
11004
11015
11106
11117
FabcM
10000
00011
10102
10113
01004
01015
01106
11117
(a) F(x,y,z)= ∑(1,4,5,6,7) (b) F(a,b,c)= ∑(0,2,3,7)
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Chapter 2
Question 7
T2T1CBA
01000
01100
01010
10110
10001
10101
10011
10111
Simplify the following Boolean function T1 and T2 to minimum number of literals:
T1= A’B’C’ + A’B’C + A’BC’
= A’B’(C’+C) + A’C’(B’+B)
= A’B’ + A’C’
= A’(B’ + C’)
T2=T1’
T2= [A’(B’ + C’)]’
= A + BC
Chapter 2
Question 8
Express the following function as sum of minterms and as product of maxterms:
F(A,B,C,D) = B’D + A’D + BD
01 01 11
F=∑m(1,3,5,7,9,11,13,15)
F=∑M(0,2,4,6,8,10,12,14)
M A B C D B’D A’D BD0 0 0 0 01 0 0 0 1 x X2 0 0 1 03 0 0 1 1 x X4 0 1 0 05 0 1 0 1 X X6 0 1 1 07 0 1 1 1 x X8 1 0 0 09 1 0 0 1 x10 1 0 1 011 1 0 1 1 x12 1 1 0 013 1 1 0 1 X14 1 1 1 015 1 1 1 1 x
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Chapter 2
Question 11
Convert each of the following expression into sum of products and product of sum:
(a)(u + xw)(x + u’v)
= ux + uu’v + xwx + xwu’v
= ux + 0 + xw + xwu’v
= ux + xw( 1 + u’v)
= ux + xw (SOP form)
= x(u + w) (POS form)
Remember
(1 + u’v)=1
Chapter 2
Question 12
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